Question

margin of error for proportions
score: 0/5 penalty: 1 off
question
a survey was given to a random sample of 650 residents of a town to determine whether they support a new plan to raise taxes in order to increase education spending. of those surveyed, 40% of the people said they were in favor of the plan. at the 95% confidence level, what is the margin of error for this survey expressed as a percentage to the nearest tenth? (do not write ±).
answer attempt 1 out of 2

Explanation:

Step1: Identify the formula for margin - of - error for proportions

The formula for the margin of error $E$ for a proportion in a large - sample case is $E = z\sqrt{\frac{p(1 - p)}{n}}$, where $z$ is the z - score corresponding to the desired confidence level, $p$ is the sample proportion, and $n$ is the sample size. For a 95% confidence level, the $z$ - score $z = 1.96$, $p=0.4$ (since 40% = 0.4), and $n = 650$.

Step2: Calculate the value inside the square - root

First, calculate $p(1 - p)$: $p(1 - p)=0.4\times(1 - 0.4)=0.4\times0.6 = 0.24$. Then, divide by the sample size $n$: $\frac{p(1 - p)}{n}=\frac{0.24}{650}\approx0.00036923$.

Step3: Calculate the square - root value

$\sqrt{\frac{p(1 - p)}{n}}=\sqrt{0.00036923}\approx0.0192$.

Step4: Calculate the margin of error

Multiply the square - root value by the $z$ - score: $E = 1.96\times0.0192\approx0.0376$.

Step5: Convert to percentage

To convert the margin of error to a percentage, multiply by 100: $E = 0.0376\times100 = 3.8\%$ (rounded to the nearest tenth).

Answer:

3.8%